LINEAR ALGEBRA
and VECTOR GEOMETRY
Volume 1 of 2
September 2014 edition
Because the book is so large,
the entire Linear Algebra course
has been split into two volumes.
© Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
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HOW TO USE THIS BOOK
I have broken the course up into lessons. Study each lesson until you can do all of my
lecture problems from start to finish without any help. Then do the Practise Problems for that
lesson. If you are able to solve all the Practise Problems I have given you, then you should have
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I have presented the course in what I consider to be the most logical order. Although my
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a different order or omit a topic. It is also possible he/she will introduce a topic I do not cover.
Make sure you are attending your class regularly!
Stay current with the
material, and be aware of what topics are on your exam. Never forget, it is your
prof that decides what will be on the exam, so pay attention.
Note that the Distance Ed course does Lesson 9 and Lesson 10 first in my
book. It then goes back to Lesson 1 and follows sequentially from there.
If you have any questions or difficulties while studying this book, or if you believe you
have found a mistake, do not hesitate to contact me. My phone number and website are noted
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I welcome your input and questions.
Wishing you much success,
Grant Skene
Owner of Grant’s Tutoring
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TABLE OF CONTENTS FOR VOLUME 1
(These Lessons are in Volume 1)
Lesson 1: Systems of Linear Equations
The Lecture ................................................................................................................................................................................................... 1
The Lecture Problems.................................................................................................................................................................................. 19
Lesson 2: Row-Reduction and Linear Systems
Lecture Problems ........................................................................................................................................................................................ 20
The Lecture ................................................................................................................................................................................................. 24
Homework and Practise Problems ............................................................................................................................................................... 72
Solutions to Practise Problems .................................................................................................................................................................... 80
Lesson 3: Matrix Math
Important Matrix Facts and Definitions. ...................................................................................................................................................... 85
Lecture Problems ........................................................................................................................................................................................ 87
The Lecture ................................................................................................................................................................................................. 88
Homework and Practise Problems ............................................................................................................................................................. 113
Solutions to Practise Problems .................................................................................................................................................................. 119
Lesson 4: The Inverse of a Matrix and Applications
Lecture Problems ...................................................................................................................................................................................... 126
The Lecture ............................................................................................................................................................................................... 128
Homework and Practise Problems ............................................................................................................................................................. 137
Solutions to Practise Problems .................................................................................................................................................................. 150
Lesson 5: Elementary Matrices
Lecture Problems ...................................................................................................................................................................................... 158
The Lecture ............................................................................................................................................................................................... 160
Homework and Practise Problems ............................................................................................................................................................. 183
Solutions to Practise Problems .................................................................................................................................................................. 185
Lesson 6: Determinants and Their Properties
Important Determinant Facts and Properties. ........................................................................................................................................... 187
Lecture Problems ...................................................................................................................................................................................... 189
The Lecture ............................................................................................................................................................................................... 191
Homework and Practise Problems ............................................................................................................................................................. 221
Solutions to Practise Problems .................................................................................................................................................................. 231
Lesson 7: The Adjoint Matrix
Lecture Problems ...................................................................................................................................................................................... 236
The Lecture ............................................................................................................................................................................................... 237
Homework and Practise Problems ............................................................................................................................................................. 254
Solutions to Practise Problems .................................................................................................................................................................. 260
Lesson 8: Cramer’s Rule
Lecture Problems ...................................................................................................................................................................................... 264
The Lecture ............................................................................................................................................................................................... 265
Homework and Practise Problems ............................................................................................................................................................. 267
Solutions to Practise Problems .................................................................................................................................................................. 271
THE MIDTERM EXAM NORMALLY COVERS
© Grant Skene for
Grant’s Tutoring
(www.grantstutoring.com)
LESSONS 1 TO 8.
DO NOT RECOPY
TABLE OF CONTENTS FOR VOLUME 2
(These Lessons are in Volume 2)
Lesson 9: Vectors
Important Vector Formulas and Facts ........................................................................................................................................................273
Lecture Problems .......................................................................................................................................................................................275
The Lecture ................................................................................................................................................................................................276
Homework and Practise Problems .............................................................................................................................................................313
Solutions to Practise Problems ..................................................................................................................................................................319
Lesson 10: Lines and Planes
Equations of Lines and Planes, etc. ............................................................................................................................................................325
Lecture Problems .......................................................................................................................................................................................326
The Lecture ................................................................................................................................................................................................327
Homework and Practise Problems .............................................................................................................................................................346
Solutions to Practise Problems ..................................................................................................................................................................352
Lesson 11: Vector Spaces and Subspaces
The Definition of Subspace and The 10 Axioms of Vector Space. ...............................................................................................................358
Lecture Problems .......................................................................................................................................................................................359
The Lecture ................................................................................................................................................................................................361
Homework and Practise Problems .............................................................................................................................................................391
Solutions to Practise Problems ..................................................................................................................................................................394
Lesson 12: Linear Independence
Key Definitions and Facts. ..........................................................................................................................................................................398
Lecture Problems .......................................................................................................................................................................................399
The Lecture ................................................................................................................................................................................................400
Homework and Practise Problems .............................................................................................................................................................424
Solutions to Practise Problems ..................................................................................................................................................................428
Lesson 13: Basis and Dimension
The Definition of Basis. ..............................................................................................................................................................................433
Lecture Problems .......................................................................................................................................................................................434
The Lecture ................................................................................................................................................................................................436
Homework and Practise Problems .............................................................................................................................................................450
Solutions to Practise Problems ..................................................................................................................................................................459
Lesson 14: Markov Analysis
Important Facts about Markov Analysis .....................................................................................................................................................467
Lecture Problems .......................................................................................................................................................................................468
The Lecture ................................................................................................................................................................................................470
Homework and Practise Problems .............................................................................................................................................................500
Solutions to Practise Problems ..................................................................................................................................................................503
Lesson 15: Linear Transformations
Important Facts and Defintions about Linear Transformations ................................................................................................................ 15-1
Lecture Problems ................................................................................................................................................................................... 15-10
The Lecture ............................................................................................................................................................................................ 15-13
Lesson 16: Eigenvalues & Eigenvectors
Important Facts and Defintions about Eigenvalues & Eigenvectors ......................................................................................................... 16-1
Lecture Problems ..................................................................................................................................................................................... 16-2
The Lecture .............................................................................................................................................................................................. 16-3
THE MIDTERM EXAM NORMALLY COVERS
© Grant Skene for
Grant’s Tutoring
(phone (204) 489-2884)
LESSONS 1 TO 8.
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
1
LESSON 1: SYSTEMS OF LINEAR EQUATIONS
Warning:
The following lesson is intended as a review of and introduction to basic
concepts of linear systems. I think you will find this lesson helpful to give you context for this
course, but it is quite possible you will never be tested on the material and methods taught here. It
will, however, give you the necessary background to understand and appreciate the later lessons.
E
A linear equation has one or more variables (like x or y) raised to the
power of 1. For example, 2x  3 y  6 is a linear equation; both x and y have understood
powers of 1. An equation is nonlinear if it has any variables raised to other powers (like x 2 or
(like
x or
3
y ); if the variables are in denominators
PL
y 3 ); if the variables are under roots (like
6
); if the variables are part of a transcendental function (examples of transcendental
x
functions are trigonometric functions like sin x ; exponential functions like e x or 2 y ;
logarithmic functions like ln x or log x ). A term that contains more than one variable
is nonlinear ( 6xy is a nonlinear term because it has two variables multiplying together even
though both of those variable are raised to the understood power of 1). The coefficients
SA
M
(the numbers in front of the variables) can come in all shapes and sizes,
however. There is also no limit on the amount of variables in a linear equation,
so long as the variables are strictly and only raised to the power of 1.
Here are some examples:

2x  3 y  4 z  7 is a linear equation. Although, there are three variables (x, y, z),
they are all raised to the power of 1, and so are linear.

3x
and “
2
y  12 is a linear equation. Even though it has weird coefficients like “ 3 ”
5
2
”, its variables are raised to the power of 1 (“x” and “y”) making it linear.
5

3 x  4 y  7 is a nonlinear equation because of the “ y ” term.

3x  4 xy  5 y  10 is a nonlinear equation because of the “xy” term.

4 x 2  5 x  4 y  8 is a nonlinear equation because of the “ x 2 ” term.

6sin x  3cos y  log 3 x  10 is a nonlinear equation. You’ve got to be kidding me!
It’s not even close; it has trigonometric and logarithmic functions in it.
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
2
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
GRAPHING A LINEAR EQUATION
The fundamental linear equation has two variables (we usually designate them by x and
y, but any symbols could be used). Linear equations are so-called because they graph
as a line. The standard form of a linear equation is ax+by = c where a, b and c are any
real number constants. For example, 2x  3 y  6 is a linear equation in standard form.
E
To quickly graph a line, we need only plot two points. The easiest points to plot are the
intercepts. To get the y-intercept, sub in x = 0. To get the x-intercept, sub in
x
0
PL
y = 0. If I wanted to graph 2x  3 y  6 , I would make a table-of-values like so:
y
sub x = 0 into 2x  3 y  6 to solve y
sub y = 0 into 2x  3 y  6 to solve x
0
Therefore, the table of values for 2x  3 y  6 would be:
SA
M
x y
0 2
3 0
We could now plot these two points and draw a line through them to make our graph.
y
2
1
2x  3 y  6
x
1 2 3
If you have two or more equations, you have a system of equations. The goal is to
then find the solution or solutions that satisfy all the equations. Geometrically speaking
(i.e. if we were looking at a graph of the system), we are trying to find the intersection
of the graphs; the point or points where the separate graphs contact each other.
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
3
LINEAR SYSTEMS WITH TWO VARIABLES
Let’s first focus on the most straightforward system of equations: two linear equations
with two variables. Geometrically speaking, we have two lines and want to find where they
intersect. There are three possibilities:
1.
The two lines are right
on top of each other;
they have infinite
points of intersection.
E
The two lines are
parallel; they do not
intersect at all.
PL
The two lines have a
single point of
intersection.
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
 2 x +3 y = 6

5 x +2 y =  7
SA
M
SOLUTION
In the elimination method we add the columns in such a way that one of
the variables is eliminated.* Essentially, the terms to be eliminated must have
identical coefficients, but with the opposite sign. We can multiply an equation by any
number we want to accomplish this (just make sure you multiply both sides of the equation to
maintain balance).
For no particular reason, I will eliminate the “y” terms (I could just as easily eliminate
the “x” terms). I will multiply every term in the first equation by −2 to create a “−6y” term and
multiply every term in the second equation by 3 to create a “+6y” term.
2 x  3 y  6  multiply by  2  4 x  6 y  12
5 x  2 y  7 
multiply by 3
 15 x  6 y  21
Add the columns 
11 x
 33
x=
 33
=3
11
*
Some people prefer to subtract the columns to eliminate a variable. I strongly advise against this as many
students often carelessly losing track of negative signs while performing the math.
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
4
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Now that we have solved x, we can substitute this value back into either one of the
original equations to solve y. I will sub it into the first equation, but I could just as easily use
the second one (either equation better produce the same value for y, or we have definitely
made a mistake).
Sub x = −3 into 2x  3 y  6 :
12
=4
3
E
2  3  3 y  6   6  3 y  6  3 y  12  y =
We have established the solution to this system is x = −3, y = 4. Put another way, we
have found both lines intersect at the point (−3, 4).
PL
We can check our answer by confirming (−3, 4) satisfies both equations.
Subbing (−3, 4) into 2x  3 y  6 , we get 2  3  3  4   6   6  12  6  6  6
Subbing (−3, 4) into 5x  2 y  7 , we get 5  3  2  4   7   15  8  7   7  7
Thus, both lines pass through the point (−3, 4).
SA
M
The solution to this system of equations is x =−3, y = 4. Interpreting
this solution geometrically, we have discovered a graph of these two lines
intersects at the point (−3, 4).
Although the question does not ask us to display the graphs, let’s do so just to visualize
what we mean by interpreting the solution geometrically. As our check confirmed, the two
lines cross at the point (−3, 4) verifying that is the one and only solution to this system of
linear equations.
y
(−3, 4)
2x  3 y  6
x
5x  2 y  7
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
2.
5
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
 x
 4 + y =1

 x  5y = 3
 6 3
E
SOLUTION
If they are nasty enough to put fractions in an equation, get rid of them! (The fractions,
not the people who put them there.) Multiply the equation by the common denominator. The
first equation has a denominator of 4, so I will multiply every term by 4 to get rid of it. The
PL
second equation has denominators of 6 and 3, so the common denominator is 6. I will multiply
every term by 6 to get rid of them.
x
 y  1  multiply by 4 
4
4x
 4y  4
4
x 5y

3
6 3
6x

6
 multiply by 6 
2

x 4y  4
6 5y
 6  3  x  10 y  18
3
SA
M
 x 4y  4
Thus, the given system of equations is equivalent to the system: 
 x  10 y  18
x  4y  4

x  10 y  18
 multiply by  1 
 x  10 y  18
Add the columns 
14 y  14
 14
y=
= 1
14
leave it alone

x  4y  4
Sub y = −1 into either one of the two equations to get x. I will use x  4 y  4 :
x  4  1  4  x  4  4  x = 8
Sub (8, −1) into both of the original equations to check the answer:
x
8
 y  1  sub in  8,  1 
  1  1  2  1  1  1  1
4
4
x 5y
8 5  1
4 5
9

 3  sub in  8,  1 

3 
 3 
3
6 3
6
3
3 3
3
The solution to this system of equations is x = 8, y = −1.
These two lines intersect at the point (8, −1).
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
6
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
3.
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
 3 x +4 y = 5

5 x  2 y =12
SOLUTION
 leave it alone 
3x  4 y  5
E
3x  4 y  5
5 x  2 y  12  multiply by 2  10 x  4 y  24
 29
29
x=
13
PL
Add the columns  13 x
Tip: Rather than go through the ordeal of fraction math to solve
y by substitution, go back to the original system and eliminate x this
time.
3x  4 y  5

multiply by 5

15 x  20 y  25
5 x  2 y  12  multiply by  3   15 x  6 y  36
SA
M
Add the columns 
26 y  11
11
y =
26
Check (29/13, −11/26) is the correct solution.*
Subbing (29/13, −11/26) into 3x  4 y  5 , we get :
87
 29 
 11 
3   4  
5 


13
 13 
 26 
22
44
87 22
65
5 

5 
5
13 13
13
13 26
Subbing (29/13, −11/26) into 5x  2 y  12 , we get :
145
 29 
 11 
5   2 
 12 


13
 13 
 26 
11
22
145 11
156
 12 

 12 
 12
13 13
13
13 26
The solution to this system of equations is x = 29/13, y = −11/26.
These two lines intersect at the point (29/13, −11/26).
*
Never check your solutions to exam questions until you have completed the entire exam. Don’t waste time
checking answers when you have other questions to do. If you’re right, you just wasted precious time
proving it; if you are wrong, you don’t want to know! Get the test finished first, then check if time allows.
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
4.
7
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
2 x +3 y = 6

4 x +6 y = 6
SOLUTION
4x  6y  6 
leave it alone
E
2 x  3 y  6  multiply by  2   4 x  6 y  12

4x  6y 
Add the columns 
6
0  6?
Whoa! What happened here? Both variables got eliminated at the same time! That
PL
left us with just “0” on the left hand side of the equation after we added the columns.
Specifically, we got “0 = −6”. This is clearly a false statement; 0 and −6 are not equal at all!
If, when performing the elimination method on a system of two
linear equations with two variables, you end up eliminating both
variables at the same time, there are two possibilities:
You end up with a false equation “0 = k” where k is a nonzero
SA
M

number. The false statement tells us there is no solution to
the system; the lines must be parallel.

You end up with the true equation “0 = 0”.
This true
statement tells us there are infinite solutions to the system;
the lines must be right on top of each other; any point on the
first line will also be on the second line.
There is no solution to this system of equations since 0 ≠ −6.
Interpreting this solution geometrically, we have discovered the two lines
are parallel and, therefore, do not intersect.
Although the question does not ask us to display the graphs, let’s do so. As we can see
on the next page, the two lines are indeed parallel.
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
8
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Table of Values
Table of Values
y
4x  6 y  6
2x  3 y  6
x
y
0 1
3/2 0
x y
0 2
3 0
2x  3 y  6
4x  6 y  6
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
PL
5.
E
x
 x  4y = 4

2 x + 8 y = 8
SOLUTION
x 4y  4
 multiply by 2 
2x  8 y  8
2 x  8 y  8
 leave it alone 
2 x  8 y  8
SA
M
Add the columns 
0  0
Since the elimination has resulted in “0 = 0”, we discover this system has infinite
solutions. In fact, we have discovered these two equations are actually multiples of each other
and, therefore, really the same line.
Table of Values
Table of Values
y
x 4y  4
2x  8 y  8
x y
0 −1
4 0
x y
0 −1
4 0
x
x 4y  4
and
2x  8 y  8
Just because there are infinite solutions does not mean everything is a solution.
Infinite does not mean everything. For example, as we can see on the graph of this
system above, (0, 0) is not a solution to this system since it is not on the lines. Only points on
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
9
the lines are solutions to this system. Admittedly, there are infinite points on the lines, but that
is nothing compared to the amount of points not on the lines.
When there are infinite solutions, we must tell people what all the solutions are. They
have to be clear which points are solutions and which are not. One way is to pick whichever of
the two equations you like (since they are describing the same line anyway), and tell them the
E
solutions are all the points on that line. So, I could say, the solution to this system is the
infinite number of points on the line x  4 y  4 . But that’s not good enough. Especially by the
time we get to Lesson 2 and encounter larger, more complicated systems of linear equations, we
need a more thorough way of describing the infinite solutions.
PL
We introduce a parameter and state all the variables in terms of it.
A
parameter is a free variable, free to be any real number. The most common letter
we use to represent a parameter is t; another commonly used symbol is s, but you could really
use any letter you want. This problem has two variables, x and y. We can pick whichever one
we want and simply let it equal t. I will let y = t, which is to say, y can be any real number; y
has infinite values. (I could just as easily let x = t.) We know all the solutions satisfy the
equation x  4 y  4 .
SA
M
Sub y = t into x  4 y  4 and solve for x: x  4t  4  x = 4+4t
We now have a “recipe” for all the solutions to the system: x = 4 + 4t, y = t. Any real
number we choose for t will produce a solution to the system. For example, if we let t = 0, we
get x = 4, y = 0. If we let t = 3, we get x = 16, y = 3. There are infinite choices for t (we
could let t = −7, t = 1/3, t =
5 , any real number we can think of), producing infinite
solutions to this system.
There are infinite solutions to this system of equations since 0 = 0.
The solutions are x = 4 + 4t, y = t where t is any real number.
Interpreting this solution geometrically, we have discovered the two lines
are, in fact, the same line. All points in the form (4 + 4t, t) are solutions
to this system.*
If you let x = s instead (I could have used t again, but I don’t want this answer to be confused with the answer
4s
4
s
s
above), and sub that into x  4 y  4 , we get s  4 y  4   4 y  4  s  y 


 y  1 
4
4 4
4
s
s

Thus, x = s, y  1  or  s,  1   is an equivalent answer (it generates all the same points).
4
4

*
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
10
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
LINEAR SYSTEMS WITH THREE VARIABLES
If you have a linear equation with three variables, ax  by  cz  d , you actually have a
plane rather than a line. For example, x  2 y  3z  6 is a plane in standard form. A plane is
a flat, two-dimensional surface; i.e. it has length and width. A table-top is a plane; the floor is a
plane; the walls are planes; the slanted roof on the outside of a typical home is a plane. The
E
equation of a plane is still considered a linear equation because all its variables are raised to the
power of 1.*
We are now dealing with three-dimensional coordinate geometry. Assuming you are in a
PL
nice ordinary rectangular room right now, take a look at a corner on the floor. Visualize the xaxis and y-axis starting at that corner and running along the edges of the floor. Say the x-axis
runs along the north-south edge of the floor, and the y-axis runs along the east-west edge of the
floor (you don’t need a compass; decide for yourself what is north, west, east, and south).
Now, in that same corner where the x-axis and y-axis started, the vertical line running up from
the floor to the ceiling is the z-axis; i.e. the z-axis is that seam where the “north” wall and the
“west” wall meet.
SA
M
Essentially, up to now, we have been restricted to drawing graphs on the floor, the xy-
plane. With the addition of the z-variable, we can now rise up off the floor into the third
dimension.
Don’t worry!
This is not a course about trying to draw three-
dimensional graphs. But, it might help to try to visualize what we are dealing with here.
Just as we do for lines, we can graph a plane by plotting the intercepts. Since we are
dealing with three variables, x, y, z, set two of them equal to 0 and sub in to the plane equation
to compute the remaining variable’s intercept.
The table of values for x  2 y  3z  6 would be:
x
0
0
6
y
0
3
0
z
2
0
0
We could now plot these three points and connect the dots to form a triangle. That
triangle becomes the base we can rest the entire plane on. Again, look at that corner of the
*
By the way, a linear equation with 4 variables or more is called a hyperplane. This is impossible for the
ordinary person to visualize since we are dealing with four dimensions or more in space.
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
11
floor where you are visualizing the three axes. The point (6, 0, 0) in our table above tells us to
go 6 units along the x-axis and plot a point there (let’s say we go 6 inches along our north-south
edge); (0, 3, 0) tells us to plot a point 3 units along the y-axis (3 inches along our east-west
line); (0, 0 , 2) plots a point 2 units up the z-axis (2 inches up the seam where the “north” and
“west” walls meet. If you want, pull out a tape measure and actually try marking those points
E
on the floor and walls (if you don’t have a life, I mean). If you were to connect those three dots
with some string, you have formed the triangular base that supports the plane. Note, the plane
would be making an angle with the floor and walls; it is not parallel to any of them.
Below is how we would attempt to depict this on paper. Note that we only draw the
PL
triangle connecting the three intercepts, but it is understood the plane is extending infinitely in
all directions from this triangular base it rests upon. Understand we are trying to show three
dimensions on two-dimensional paper, so always try to hold on to the image of the walls and
floor to properly see this.
z
SA
M
2
1
3
x  2 y  3z  6
y
1 2 3
6
x
Let me stress, this is not a course about drawing graphs in three-
dimensional space. I am merely doing this as an exercise, so that
you might grasp visually what we are dealing with. It is unlikely you
will have to draw a graph like this on your exam (it has happened
once or twice though, so never say never).
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
12
6.
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
 x +2 y + 3 z = 6

2 x +3 y + 5 z = 9
5 x +2 y  z =  2

SOLUTION
equations are involved.
E
There is a two-stage process to the elimination method when three
It helps to keep track of things if we number the
original equations (1), (2) and (3).
PL
Stage 1: Select a pair of equations and eliminate whichever variable
strikes your fancy to create an equation that has only two
variables. Number that new equation (4). Then, select a second
pair of equations and eliminate the same variable.
This is a
must! If you eliminated x in the first pair, you must eliminate x
in the second pair. Number that new equation (5).
Stage 2: Now equations (4) and (5) form a system of equations with two
Solve that system by elimination.
SA
M
variables.
Once you have
solved those two variables, you can sub them into any one of (1),
(2) or (3) to solve the remaining variable.
Number the original equations:
(1) x  2 y  3z  6

(2) 2 x  3 y  5z  9
(3) 5 x  2 y  z   2

I like that “–z” term in equation (3), so I will exploit it to eliminate the “z” terms in my
pairs. (Another good choice would be to exploit the “x” term in equation (1) to eliminate the
“x” terms in the pairs.)
My first pair will be equations (1) and (3):
(1)
x  2 y  3z  6  leave it alone 
x  2 y  3z  6
(3) 5 x  2 y  z   2  multiply by 3  15 x  6 y  3z  6
Add the columns  16 x  8 y
 0
Equation (4)
16 x +8 y = 0
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
13
My second pair will be equations (2) and (3):
(2) 2 x  3 y  5z  9  leave it alone 
2 x  3 y  5z 
9
(3) 5 x  2 y  z   2  multiply by 5  25 x  10 y  5z  10
Add the columns  27 x  13 y
 1
Equation (5)
27 x +13 y = 1
E
Equations (4) and (5) now form a system of two equations with two
variables:
(4) 16 x  8 y  0

(5) 27 x  13 y  1
PL
Here’s a good idea: Divide equation (4) by 8 to make the coefficients smaller and easier
to work with. (Note: 0  8 = 0.)
(4) 16 x  8 y  0
 divide by 8  2 x  y  0
(5) 27 x  13 y  1  leave it alone  27 x  13 y  1
Now I will eliminate y from this system:
2x  y  0
 multiply by  13  26 x  13 y  0
27 x  13 y  1 
leave it alone

SA
M
Add the columns 
27 x  13 y  1
x
 1
x = 1
Sub x = −1 into 2x  y  0 :
2  1  y  0   2  y  0  y =2
We have established so far x = −1, y = 2. Sub these into any one of the original three
equations to solve z. I will use equation (3) 5x  2 y  z  2 :
5  1  2  2  z  2   5  4  z  2   1  z  2   z  1  z =1
Thus, x = −1, y = 2, z = 1 or (–1, 2, 1) is the solution to this system. By the way,
don’t get confused and say this system has three solutions; this system has one solution. That
one solution contains values for all three variables.
If time allows, we can check our answer by confirming (−1, 2, 1) satisfies all three of the
original equations in the system. If the check fails in any single one of the equations, we have
made a mistake.
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
14
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Subbing (−1, 2, 1) into equation (1) x  2 y  3z  6 , we get:
1  2  2  3 1  6   1  4  3  6  6  6
Subbing (−1, 2, 1) into equation (2) 2x  3 y  5z  9 , we get:
2  1  3  2  5 1  9   2  6  5  9  9  9
E
Subbing (−1, 2, 1) into equation (3) 5x  2 y  z  2 , we get:
5  1  2  2  1  2   5  4  1  2   2  2
Thus, all three planes pass through the point (−1, 2, 1).
PL
The solution to this system of equations is x =−1, y = 2, z = 1.
Interpreting this solution geometrically, we have discovered a graph of
these three planes intersects at the point (−1, 2, 1).
Don’t even think about trying to draw a graph of these three planes to visualize them
intersecting at this one point. It isn’t worth the effort, and your picture is probably going to
look like somebody spilled the uncooked spaghetti.
SA
M
Here is a way to get a grasp of this visually. Look at the “north” wall of your room.
That’s sort of like plane (1). Now look at the “west” wall of your room. That’s sort of like plane
(2). Note these two planes intersect along the infinite number of points on the line running up
the seam where the two walls meet (that seam in the “northwest” corner running from the floor
up to the ceiling). Admittedly, these two walls make a right angle with each other, while the
two planes in our system may make some other angle, but who cares? Visualize swinging the
two walls using that “northwest” seam as a hinge, like swinging the covers of a textbook. The
planes can make any angle you want, but they still intersect along that line running up the
seam. Finally, look at the floor. That’s sort of like plane (3). Note the floor shares a seam with
the “north” wall (infinite points along their line of intersection). The floor also shares a seam
with the “west” wall (infinite points along their line of intersection). But, there is only one
point where the floor meets both the “north” and “west” walls, and that is that point in the
corner of the floor at the “northwest” seam.
The three planes have a single point of
intersection, just as our three planes meet at the point (–1, 2, 1). (–1, 2, 1) is sort of like
that corner where the floor meets both the north wall and the south wall.
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
7.
15
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
2 x  y  4 z = 0

 x +2 y + 3 z = 1
2 x + y + 5 z = 2

SOLUTION
E
Number the original equations:
PL
(1) 2 x  y  4 z  0

(2) x  2 y  3z  1
(3) 2 x  y  5z  2

I like that “–y” term in equation (1), so I will exploit it to eliminate the “y” terms in my
pairs. (Another good choice would be to exploit the “x” term in equation (2) to eliminate the
“x” terms in the pairs.)
My first pair will be equations (1) and (2):
(1) 2 x  y  4 z  0  multiply by 2  4 x  2 y  8 z  0
(2)
x  2 y  3z  1  leave it alone 
x  2 y  3z  1
SA
M
Add the columns  5 x
 5z  1
Equation (4)
5 x  5 z =1
My second pair will be equations (1) and (3):
(1) 2 x  y  4 z  0  leave it alone  2 x  y  4 z  0
(3) 2 x  y  5z  2  leave it alone  2 x  y  5z  2
Add the columns  4 x
z2
Equation (5)
4x+ z =2
We now have a system of two equations with two variables:
(4) 5 x  5z  1

(5) 4 x  z  2
I will eliminate z from this system:
(4) 5 x  5z  1
(5)
4x  z  2
 leave it alone 
5 x  5z  1
 multiply by 5  20 x  5z  10
Add the columns  25 x
 11
11
x=
25
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
16
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Since x = 11/25 is too annoying to sub into one of equations (4) or (5) to
solve z, I will perform elimination again; this time eliminating x:
(4) 5 x  5z  1  multiply by  4   20 x  20 z  4
(5)
4x  z  2 
multiply by 5

20 x
25z  6
6
z=
25
E
Add the columns 
 5z  10
We have established so far x = 11/25, z = 6/25. Sub these into any one of the
original three equations to solve y. I will use equation (1) 2x  y  4 z  0 :
PL
22
24
2
2
2
 11 
 6 
2   y  4    0 
 y
0   y
0   y
 y =
25
25
25
25
25
 25 
 25 
Thus, x = 11/25, y = –2/25, z = 6/25 or (11/25, –2/25, 6/25) is the solution to
this system.
If time allows, we can check our answer by confirming (11/25, –2/25, 6/25) satisfies all
three of the original equations in the system.
If the check fails in any single one of the
SA
M
equations, we have made a mistake. I will leave you to perform the check yourself.
11
2
6
, y =
, z=
.
25
25
25
Interpreting this solution geometrically, we have discovered a graph of
The solution to this system of equations is x =
2 6 
 11
,
,
these three planes intersects at the point 
.
25 25 
 25
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
8.
17
Solve the system of equations below using the elimination method, and
interpret the solution geometrically.
 2 x  y + 3z = 3

 3 x + 2 y  z = 8
SOLUTION
E
Wait a minute! There are only two equations here! All we can do then is eliminate one
of the variables. I like that “–y” term in equation (1), so I will exploit it to eliminate the “y”
terms. (Another good choice would be to exploit the “–z” term in equation (2) to eliminate the
“z” terms.)
2 x  y  3z  3  multiply by 2 
4 x  2 y  6z  6
PL
(1)
(2) 3 x  2 y  z  8  leave it alone  3 x  2 y  z  8
Add the columns 
Equation (3)
x
 5z  14
x +5 z =14
That’s as far as we can go. The solution to this system of two equations is the equation
x  5z  14 . Note: this is a linear equation with two variables in it. That means it graphs as a
line! This makes perfect sense. The original system was two planes, and we have discovered
SA
M
these planes have a line of intersection. Again, just like the “north” wall and the “west”
wall intersect along that seam running up the northwest corner of your room, two planes can
intersect along an infinite line.
There are infinite points of intersection between
these two planes, all of them lying on the line x +5z =14 .
Just as we did in question 5 above, whenever we have infinite solutions to a system of
equations, we will introduce a parameter. The easiest thing here is to make z the parameter
(but you could make x the parameter if you prefer).
I will let z = t, a parameter. Subbing z = t into x  5z  14 , we get:
x  5t  14  x =14  5t
We have established so far x = 14 – 5t, z = t. Sub these into either one of the original
two equations to solve y. I will use equation (1) 2x  y  3z  3 :
2 14  5t   y  3  t   3  28  10t  y  3t  3  28  7t  y  3
Move everything over to the right side of the equation except the “–y” term:
 y  3  28  7t   y  25  7t  multiply both sides by  1  y = 25  7t
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
18
LESSON 1: SYSTEMS of LINEAR EQUATIONS (Linear Algebra)
Thus, x = 14 – 5t, y = 25 – 7t, z = t or (14 – 5t, 25 – 7t, t) is the solution to this
system. We have given people a “recipe” to generate the infinite number of points that satisfy
this system of equations. By selecting different values of the parameter t, we generate different
solutions. For example, if t = 0, we get the solution (14, 25, 0); if t = 1, we get the solution
(9, 18, 1); if t = 2, we get (4, 11, 2); etc.
E
That is the beauty of using parameters to describe infinite solutions: we get an easy
recipe to generate all the solutions. We can let t be any real number. (The parameter t doesn’t
have to be just counting numbers like 0, 1, 2, …; we can let t be 1/3,
they all generate solutions to the system.)
5 , –4.72, whatever, and
of the equations.
PL
Let’s prove (14 – 5t, 25 – 7t, t) is the solution to the system by showing it satisfies both
Subbing (14 – 5t, 25 – 7t, t) into equation (1) 2x  y  3z  3 , we get:
2 14  5t    25  7t   3  t   3  28  10t  25  7t  3t  3  3  3
Note, the t terms cancel out.
Subbing (14 – 5t, 25 – 7t, t) into equation (2) 3x  2 y  z  8 , we get:
3 14  5t   2  25  7t    t   8   42  15t  50  14t  t  8  8  8
SA
M
Note, the t terms cancel out.
The solution to this system of equations is x = 14 – 5t, y = 25 – 7t, z = t
where t is any real number.* Interpreting this solution geometrically, we
have discovered a graph of these two planes has a line of intersection. All
the points in the form (14 – 5t, 25 – 7t, t) are on this line.
Be sure to point out that t is any real number. It is generally taken for granted that t, being a parameter, is any
real number, but some profs will deduct marks if you don’t specifically say this in your answer.
*
© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) LESSON 1: SYSTEMS of LINEAR EQUATIONS
19
LECTURE PROBLEMS
For your convenience, here are the 8 questions I used as examples in this
lesson. Do not make any marks or notes on these questions below. You want to keep these
questions untouched, so that you can look back at them without any hints. Instead, make any
necessary notes, highlights, etc. in the lecture in Lesson 1 above.
E
For questions 1 to 8 below, solve the system using the elimination method, and
interpret the solution geometrically.
 x
 4  y  1
2. 
 x  5y  3
 6 3
 3x  4 y  5
3. 
5 x  2 y  12
SA
M
2x  3 y  6
4. 
4 x  6 y  6
PL
 2x  3 y  6
1. 
5 x  2 y   7
 x 4y  4
5. 
2 x  8 y  8
 x  2 y  3z  6

6.  2 x  3 y  5z  9
5 x  2 y  z   2

2 x  y  4 z  0

7.  x  2 y  3z  1
 2 x  y  5z  2

 2 x  y  3z  3
8. 
3 x  2 y  z  8
© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
20
LESSON 2: ROW-REDUCTION AND LINEAR SYSTEMS (Linear Algebra)
Lesson 2: Row-Reduction and Linear Systems
The Rank of a Matrix:
E
} The rank of a matrix equals the number of leading 1’s it would have in its rowreduced echelon form.
} If a system is consistent (one or infinite solutions), the rank of the coefficient matrix
is equal to the rank of the augmented matrix.
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} If a system is inconsistent, the rank of the coefficient matrix is less than the rank of
the augmented matrix. (The augmented matrix will have a rank that is one higher
than the coefficient matrix.)
} The rank of the coefficient matrix could never be more than the rank of the
augmented matrix.
Lecture Problems:
(Each of the questions below will be discussed and solved in the lecture that follows.)
Suppose that the following matrices are the row echelon form of the augmented matrix
of a system of linear equations. For each matrix answer the following questions:
SA
M
1.
(i) How many equations and how many variables were in the original system?
(ii) What is the rank of the coefficient matrix and the augmented matrix?
(iii) How many parameters are in the solution?
(iv) List the solution(s), if possible.
 1 0 0 3


(a)  0 1 0 5 
 0 0 1 2 


1 2 0 3 0 4


(b)  0 0 1 2 0 3 
 0 0 0 0 1 0


 1 0 2 0 3


(c)  0 1 3 4 5 
 0 0 0 0 1


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2.
2x
x
Consider the system
21
 3y 
z  a

z  b.
y  2z  c
Suppose (1, 2, –1) is a solution to this system, find a, b and c.
2 x1
 x1
(a)
x1
3x
(b) 2 x
x

x3
 2x3
 2x3
x3
 3x4

x4
2x
(d)
x
4.




x5
x5
x5
x5




1
1
0
1
 2
 5
 2

1
 7 y  2z  9
 4 y  2z  4
 3y 
z  4

x2
 3x2
 2x2

x2

x3
 3x3
 2x3

x3

x4
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x1
2 x1
(c)
 x1
 2x2

x2

x2
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Solve the following systems of equations using Gauss-Jordan elimination.
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3.


x4
 2x4
y  z  2
y  z  1
 z  1
Solve the system of equations
x
x


y  z  3
y  z  0
 z  3
using Gaussian elimination and back substitution.
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Solve the two systems of equations below simultaneously:
x
x
6.
 6 y  3z  34
 6 y  2z  30
2 y  2z  14
and
x
x
 6 y  3z  30
 6 y  2z  24
2 y  2z  16
Given the system of equations
x1
x2
x2
 2x2
 2x3

x3
 3x3
 0
 k,
 1
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 x1

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5.
LESSON 2: ROW-REDUCTION AND LINEAR SYSTEMS (Linear Algebra)
find, if possible, the value of k if
(a) the system has infinite solutions.
(b) the system has a unique solution.
(c) the system has no solution.
Given the augmented matrix
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7.
 1 0 1 1


 0 1 1 2 ,
 0 2 a b


find conditions on real numbers a and b such that:
(a) the system has no solution.
(b) the system has a unique solution.
(c) the system has infinitely many solutions.
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8.
23
A linear system of equations has been row-reduced into this augmented matrix (it is not
necessarily in RREF)
7 
1 0 a 1


5
6 ,
0 1
 0 0 a2  4a a  4 


(a) the system has infinitely many solutions.
(b) the system has no solution.
(c) the system has a unique solution.
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find all real numbers a such that:
Anne, Betty and Carol went to their local produce store to purchase some fruit. Anne
bought one pound of apples and two pounds of bananas and paid $1.85. Betty bought
two pounds of apples and one pound of grapes and paid $3.65. Carol bought one pound
of bananas and two pounds of grapes and paid $3.95. Find the price per pound for each
of the three fruits.
10.
A company owns three types of trucks. These trucks are equipped to haul two different
types of machines per load. Truck 1 can haul 2 of machine A and 3 of machine B. Truck
2 can haul 1 of machine A and 2 of machine B. Truck 3 can haul 3 of machine A and 4
of machine B. Assuming each truck is fully loaded, how many trucks of each type should
be sent to haul exactly 18 of machine A and 26 of machine B. If there is more than one
possible solution provide all possible solutions, keeping in mind that the company can
use no more than 4 of any particular type of truck.
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9.
11.
List all 3 × 2 row-reduced echelon form matrices.
12.
Consider the linear equation with three variables:
ax  by  cz  d
1
ax  by  cz  0
 2 .
where a, b, c, and d are any real number but d ≠ 0.
Then, the associated homogeneous equation would be:
Let  x1 , y1 , z1  and  x 2 , y2 , z2  be two solutions to equation (1), and let  x 0 , y0 , z0  be
a solution to equation (2).
(a) Show  x1  x2 , y1  y2 , z1  z2  is a solution to equation (2).
(b) Show  x1  x0 , y1  y0 , z1  z0  is a solution to equation (1).
(c) Show  kx 0 , ky0 , kz0  is a solution to equation (2) where k is any real number.
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ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
Homework:
 Memorize the facts about The Rank of a Matrix on page 20.
 Study the lesson thoroughly until you can do all of questions 1 to 12 on pages 20
to 23 from start to finish without any assistance.
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 Do all of the Practise Problems below (solutions are on pages 80 to 84).
Practise Problems:
In each of the following parts (a) – (c) of this question, you are given the row-reduced
echelon form of the augmented matrix of a system of linear equations. In each case, say
whether the system is inconsistent, has a unique solution, or has infinitely many
solutions. If the system is inconsistent, explain why. If the system is consistent, state the
solution(s).
1

0
(b)  0

0
0

0
1
0
0
0
0 2

0 5 
1 6

0
1
0 0 
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1 0 0 0
(a) 

 0 0 1 3
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1.
2.
1

0
(c) 
0

0
5
0
0
0
0
1
0
0
0
0
1
0
0

0
0

0
(a) Write the augmented matrix for the system:
x1
2 x1
 2x2
 4 x2

x3
x3
 2x4

x4
 2x4
 3

1
 4
(b) Find the reduced row-echelon form for the augmented matrix in (a).
(c) Write all solutions to the system in (a).
3.
A system of linear equations is given by
x
x
3x
x

y  3z  1

y  2z  0
.
 3 y  5z 
1

y
 2
(a) Find the reduced row echelon form of the augmented matrix of the system.
(b) Write all the solutions to the system.
(c) Find the solution set which has x = 3.
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(Linear Algebra) ROW-REDUCTION PRACTISE PROBLEMS
The augmented matrix of a system of linear equations has the following row-reduced
echelon form matrix:
1
0

R  0

0
0
0
1
0
0
0
1
1
0
0
0
0
0
1
0
0
0
0
0

1
0
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4.
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(a) How many equations are there in the original system of linear equations?
(b) How many variables does the system of linear equations contain?
(c) How many solutions does the system have? Explain.
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(d) How many leading 1's does R contain?
(e) What is the rank of the coefficient matrix?
5.
Determine the number of solutions the following linear system of homogeneous
equations has without solving the system. Give a reason for your answer.
3 x1
x1
 7 x3

x3

x4
 0
 0
Each matrix A and B below is the augmented matrix of a system of linear equations in
x1, x2, x3, and x4. For each of A and B do the following.
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6.
 5x2
 3x2
(i) Put the matrix into reduced row echelon form, while stating exactly which
elementary row operations you are using.
(ii) State how many solutions the system has: none, one, or infinitely many.
(iii) If the system has solutions, give the general solution in vector form, using
parameters s, t, u, v,... (as necessary).
 1 2 1 3 2


(a) A  0 0 1 1 4 
 2 4 2 6 4 
 1 2 2 3 5
(b) B  0 1 0 1 4 
 2 4 4 6
5
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7.
ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
Given the system
2 x1
2 x1
4 x1
 4 x2
 4 x2
 8 x2
 2x3
 3x3
 3x3

x3
 2x4
 3x4
 4 x4

x4




2
2
.
5
0
(a) Solve the system of equations above by completely reducing the augmented matrix
to row-reduced echelon form.
(c) What is the rank of the augmented matrix?
Solve the following systems of equations using Gauss-Jordan or Gaussian elimination:
x1
(a)
3 x1

x3
 2x3
x1
(b) 4 x1
2 x1
 4
 7

y  z  1

y  5z  5
 3 y  5z  5
(e)
(f)
x1 
x2
2 x1  2 x 2

x3
 3x3
 2x2

x2

x2

x3
 2x3
 2x3
x3
2 x1
 x1
x1
x1
2 x1
(g)
 x1
x
(h) 2 x
4x
 2x2
 4 x2
 2x2
x2

x2

x2
 2x2

x3
 5x3
 3x3
 7
 4
 0
x  2 y  3z  1
y  3z  1
(d) 2 x 
x 
y  2z  1
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x
(c)
2x
 2x2
 4 x2
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8.
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(b) Interpret the solution geometrically (e.g., as a point, line, plane, hyperplane, etc.).
 2x3
 3x3

x4
 3x4
 3x4

x4




2x4
4 x4
2x4
3x4

x5
 4 x5




 7
 18
0
0
0
0
 5
 12
 2
 0

y 
z  1

y  2z  5
 3y
 7
© 1997-2011 Grant Skene for
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(Linear Algebra) ROW-REDUCTION PRACTISE PROBLEMS
Solve the following systems of equations using Gauss-Jordan or Gaussian elimination:
x1
(b) x1
x1
(c)



 2x3
 2x3



x2
x2
x2
x3
x3
x3
 2x4
 5x4
 7 x4

x4
 2x4
 2x4
x 
z 
x 
y  6 
y  2 
z  4 
x 
y 
x 

2y 

x5
 2 x5
 3 x5
(f)
(g)
x1
 x1
3 x1
2 x1
5 x1
 2
 3
 2
z
x
y
y 
5
z 
3
z  8
SA
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(d) 3 x
2 x1
x1
(e)
2 x1
 4
 2
 6
E
 2x2
 4 x2
 6 x2
x1
(a) 2 x1
3 x1
PL
9.
75
 2x2
 2x4
x2
 2x3

x3
 2x3
3x2
 2x2
 4 x2
 3x3
 3x3
 6 x3
 3
 2
 8

 3x2

x2

x2
 2x2
© 1997-2011 Grant Skene for
 6 x3

x3
 5x3
 5x3

x4
 2
 3
 4

3

3

1
 8
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76
 2x2
 2x2
 2x2
x1
(h) 2 x1
3 x1
(i)
(j)
 0
 6
 4


2
 4
 1
x2

x2

x3

x3
 2x3
 3x2
 6 x2
 8 x2

x3
 4 x3
 4 x3






x2
x2
x2
x4
 2x4

x4
 2x4
 3x4

x4
 2x4
 2x4
x3
x3
x3
 2 x5

x5
 4 x5

x5
 2 x5
 3 x5
 8
 14
 21
 2
 3
 2
SA
M
x1
(k) x1
x1

x4
 8 x4
 7 x4

x1
2 x1
x1
x1
x1
2 x1
 5x3
 8 x3
 3x3
E
(Continued) Solve the following systems of equations using Gauss-Jordan or Gaussian
elimination:
PL
9.
ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
(l)
(m)
x1
2 x1

x2

2 x1

x2

x3
x3
x3
 2x4
 4 x4

x4
 2x4

3
 7

1
 9
7 x1  3 x 2  x 3  0
7 x1  3 x 2  x 4  x 5  1
40 x  16 y  9z  1
(n) 13 x  5 y  3z  2
5x  2 y 
z  1
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
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(Linear Algebra) ROW-REDUCTION PRACTISE PROBLEMS
10.
77
Given the following augmented matrix for a system of linear equations:
0
0 2
1


 0 k  1 0 0 .
0
0
k 5 

E
For what value(s) of k, if any, are there:
(a) a unique solution?
(b) infinitely many solutions?
(c) no solution?
11.
PL
Give reasons for your answers.
Find all c such that the system below has no solutions.
x  2y  1
1
x 
2
12.
y  c
Find a value of p and q so that the system
SA
M
x
 z  1
y  z  1
py  qz  1
(a) has a unique solution.
(b) has an infinite number of solutions.
(c) has no solution.
© 1997-2011 Grant Skene for
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78
13.
ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
1

0
Let R  
0

0
system.
0
0
2

1
0
1 
be a row-echelon form of the augmented matrix of a linear
0 a2 b 

0
0
0
E
(a) What are the number of equations and the number of variables in the system?
(b) Find all of the values of a and b for which the system has a unique solution.
(c) Find all of the values of a and b for which the system has no solution.
14.
1

0
Let 
0

0
system.
0 1
1 2
0 a
0 0
PL
(d) Find all of the values of a and b for which the system has infinitely many solutions.
How many parameters are there in the solution set?
2 

3 
be a row-echelon form of the augmented matrix of a linear
0 

b  1
SA
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(a) What are the number of equations and the number of variables in the system?
(b) Find all values of a and b such that the system has no solutions.
(c) Find all of the values of a and b such that the system has a unique solution.
15.
5 a
 2 3


2 0  be the augmented matrix of a linear system.
Let A   1 1
 4 6 10 1 


(a) What are the number of equations and the number of variables in the system?
(b) Find all values of “a” for which the system has no solution.
(c) Find all of the values of “a” for which the system has a infinitely many solutions.
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
(Linear Algebra) ROW-REDUCTION PRACTISE PROBLEMS
16.
1 0 0

Let the augmented matrix of a linear system be given by  0 1 0
0 0 x

79
2

3.
y 
For what values of x and y is there
(a) No solution?
(c) Infinitely many solutions?
17.
E
(b) Exactly one solution?
Suppose that the augmented matrix of a linear system of equations is given by
PL
9 
 1 2 3
 1 3 0
5  .

 2 5 k 3k  5
For what values of k is there
(a) exactly one solution?
SA
M
(b) infinitely many solutions?
18.
 x  y  2z  a
Consider the system 
.
2 x  by  4 z  1
In each case below, determine all values of a and b which give the indicated number of
solutions. If no values of a and b exist, explain why not.
(a) No solution.
(b) Exactly one solution.
(c) Infinitely many solutions.
© 1997-2011 Grant Skene for
Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
SOLUTIONS TO ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
SA
M
PL
E
80
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
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81
SA
M
PL
E
(Linear Algebra) SOLUTIONS TO ROW-REDUCTION PRACTISE PROBLEMS
© 1997-2011 Grant Skene for
Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
SOLUTIONS TO ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
SA
M
PL
E
82
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
DO NOT RECOPY
83
SA
M
PL
E
(Linear Algebra) SOLUTIONS TO ROW-REDUCTION PRACTISE PROBLEMS
© 1997-2011 Grant Skene for
Grant’s Tutoring (www.grantstutoring.com)
DO NOT RECOPY
SOLUTIONS TO ROW-REDUCTION PRACTISE PROBLEMS (Linear Algebra)
SA
M
PL
E
84
© 1997-2011 Grant Skene for
Grant’s Tutoring (text or call (204) 489-2884)
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